Question
Sun September 11, 2011 By:

# What is the technique to solving questions on limiting reagent. Explain with an example.

Mon September 12, 2011

Step One: Balance the equation.
Example: 4Al(s) + 3O2(g)  2Al2O3(s)

Step Two: Convert reactants to moles.
If the reactants are given as a mass, then divide by the molecular/atomic mass of the compound/element in order to convert it to moles.
If the reactants are given as a concentration, multiply the concentration by the volume to convert it to moles.

Example: Atomic mass of Al: 29.9815 g/mol (from Periodic Table)
Molecular mass of O2: 15.9994 (from Periodic Table) x 2 = 31.9988 g/mol
So....
10.0g/26.9815g/mol = 0.3706 moles of Al
10.0g/31.9988g/mol = 0.3125 moles of O2

Step Three: Divide the moles of each reagent by its coefficient.
The coefficients are found in the balanced equation. Example: For the aluminum: 0.3706 mol/4 = 0.09265mol

Step Four: Identify the Limiting Reagent.

The limiting reagent is the smallest number. Example: In this case, 0.09265 is smaller than 0.1042, so aluminum is the limiting reagent.

Step Five: Multiply limiting reagent by coefficient of product.
Use the value from step three, not step two (i.e. the reactant coefficient must have been taken into account). Remember also to use the value for the limiting reagent only. All other reactants can be ignored.

Example: 0.09265 mol X 2 = 0.1853 mol of Al2O3

Step Six: Convert moles of product back into grams/concentration.
If the product is aqueous, you should convert your answer back to concentration by dividing by the total volume (in liters). If your product is not aqueous, you should convert to mass by multiplying by the molecular/atomic mass.

Example: Molecular mass of aluminum oxide: (2 x 29.9815) + (3 x 15.9994) = 107.9612 g/mol (all from periodic table)
0.1853 mol x 107.9612g/mol = 18.9g of aluminum oxide

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